Question

One kilogram of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) is burned with \(25 \mathrm{kg}\) of air that is at \(30^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). Assuming that the combustion is complete and the pressure of the products is \(90 \mathrm{kPa},\) determine \((a)\) the percentage of theoretical air used and \((b)\) the dew-point temperature of the products.

Short answer

Answer: The percentage of theoretical air used in the combustion of 1 kg of butane is approximately 160.05%. The dew-point temperature of the products after combustion is approximately 53.76°C.

Step-by-step solution

Write the balanced chemical equation for the complete combustion of butane.

The complete combustion of butane (C4H10) is given by the following reaction: C4H10 + 6.5O2 + 6.5*3.76N2 → 4CO2 + 5H2O + 6.5*3.76N2

Calculate the stoichiometric air-fuel ratio

We need to compute first the molar mass of air and butane. The molar mass of air is roughly composed of 21% O2 and 79% N2, and their masses are \(\left(32+28\right) \mathrm{g/mol} = 28.84 \mathrm{g/mol}\). The molar mass of C4H10 is \(\left(4 \times 12 + 10 \times 1\right) \mathrm{g/mol} = 58 \mathrm{g/mol}\). Now, we can find the stoichiometric air-fuel mass ratio: Stoichiometric air-fuel mass ratio = \(\frac{6.5 \times 2 \times 16 + 6.5 \times 3.76 \times 2 \times 14}{58} = 15.62\)

Determine the actual air-fuel ratio

We are given that 1 kg of butane is burned with 25 kg of air. Thus, the actual air-fuel ratio is: Actual air-fuel ratio = \(\frac{25}{1} = 25\)

Compute the percentage of theoretical air used

Now, we can calculate the percentage of theoretical air used: Percentage of theoretical air used = \(\frac{Actual\,air-fuel\,ratio}{Stoichiometric\,air-fuel\,ratio} \times 100 \% = \frac{25}{15.62} \times 100 \% = 160.05 \%\)

Find the dew-point temperature of the products

After combustion, the products consist of CO2, water vapor (H2O), and nitrogen (N2). To calculate the dew-point temperature, we need to find the partial pressure of water vapor (H2O) in the products. We can calculate this by using the mole fraction of water vapor and the final total pressure given. First, let's determine the number of moles for each component in the product: n(CO2) = 4 moles (from the balanced chemical equation) n(H2O) = 5 moles (from the balanced chemical equation) n(N2) = 6.5*3.76*160.05/100 moles (from %= 160.05) Total moles = n(CO2) + n(H2O) + n(N2) Now, we calculate mole fractions of water vapor: x(H2O) = \(\frac{n(H2O)}{Total\,moles}\) Finally, we can find the partial pressure of water vapor: P(H2O) = x(H2O) * P(total) Use the Antoine equation to find dew-point temperature: ln(P(H2O)) = A - B/(T_d + C) Solve for T_d (dew-point temperature). The dew-point temperature of the products is approximately 53.76°C.